Algorithms/JavaAlgorithms
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Fixed Compiler warnings, closed the scanned and fixed some typos

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This commit is contained in:
rmakynen 2018-10-15 09:21:30 +03:00
parent c75fce171b
commit c6e6610909
1 changed files with 15 additions and 15 deletions
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30
Others/Huffman.java
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@ -1,4 +1,3 @@
import java.util.Comparator;
import java.util.Iterator;
import java.util.LinkedList;
@ -14,24 +13,24 @@ Output :
Enter number of distinct letters
6
Enter letters with its frequncy to encode
Enter letters with its frequency to encode
Enter letter : a
Enter frequncy : 45
Enter frequency : 45
Enter letter : b
Enter frequncy : 13
Enter frequency : 13
Enter letter : c
Enter frequncy : 12
Enter frequency : 12
Enter letter : d
Enter frequncy : 16
Enter frequency : 16
Enter letter : e
Enter frequncy : 9
Enter frequency : 9
Enter letter : f
Enter frequncy : 5
Enter frequency : 5
Letter Encoded Form
a 0
@ -64,17 +63,17 @@ public class Huffman {
// A simple function to print a given list
//I just made it for debugging
public static void print_list(List li){
public static void print_list(List<Node> li){
Iterator<Node> it=li.iterator();
while(it.hasNext()){Node n=it.next();System.out.print(n.freq+" ");}System.out.println();
}
//Function for making tree (Huffman Tree)
public static Node make_huffmann_tree(List li){
public static Node make_huffmann_tree(List<Node> li){
//Sorting list in increasing order of its letter frequency
li.sort(new comp());
Node temp=null;
Iterator it=li.iterator();
Iterator<Node> it=li.iterator();
//System.out.println(li.size());
//Loop for making huffman tree till only single node remains in list
while(true){
@ -89,7 +88,7 @@ public class Huffman {
//Below condition is to check either list has 2nd node or not to combine
//If this condition will be false, then it means construction of huffman tree is completed
if(it.hasNext()){b=(Node)it.next();}
//Combining first two smallest nodes in list to make its parent whose frequncy
//Combining first two smallest nodes in list to make its parent whose frequency
//will be equals to sum of frequency of these two nodes
if(b!=null){
temp.freq=a.freq+b.freq;a.data=0;b.data=1;//assigining 0 and 1 to left and right nodes
@ -109,7 +108,7 @@ public class Huffman {
//Function for finding path between root and given letter ch
public static void dfs(Node n,String ch){
Stack<Node> st=new Stack(); // stack for storing path
Stack<Node> st=new Stack<Node>(); // stack for storing path
int freq=n.freq; // recording root freq to avoid it adding in path encoding
find_path_and_encode(st,n,ch,freq);
}
@ -140,15 +139,16 @@ public class Huffman {
System.out.println("Enter number of distinct letters ");
int n=in.nextInt();
String s[]=new String[n];
System.out.print("Enter letters with its frequncy to encode\n");
System.out.print("Enter letters with its frequency to encode\n");
for(int i=0;i<n;i++){
Node a=new Node();
System.out.print("Enter letter : ");
a.letr=in.next();s[i]=a.letr;
System.out.print("Enter frequncy : ");
System.out.print("Enter frequency : ");
a.freq=in.nextInt();System.out.println();
li.add(a);
}
in.close();
Node root=new Node();
root=make_huffmann_tree(li);
System.out.println("Letter\t\tEncoded Form");